Show that $\lim_{n\to\infty}\int_{X}fd\mu_n = \int_Xfd\mu$

convergence-divergencemeasure-theoryreal-analysis

Given is a a $\sigma$ – Algebra $\mathcal{A}$, a measure space $(X, \mathcal{A})$ and a sequence of measures $\left(\mu_n :\mathcal{A} \to [0,\infty]\right)_{n\in \mathbb{N}}$ with $\mu_n(A) \leq \mu_{n+1}(A)\:\forall n\in \mathbb{N}$. I have to show that if $f:X\to[0,\infty]$ is measureable with respect to $\mathcal{A}$, it follows that $\lim_{n\to\infty}\int_{X}fd\mu_n = \int_Xfd\mu$, where $\mu: \mathcal{A}\to[0,\infty]$, $\mu(A):=\sup\{\mu_n(A):n\in\mathbb{N}\}$ and $A\in\mathcal{A}$. We have defind integrals with step functions, hence $f=\sum_{k=1}^N\alpha_k1_{A_k}$ and it follows by definition that $\int_Xfd\mu=\sum_{k=1}^N\alpha_k\mu(A_k)$ for an arbitrary measure $\mu$. I have literally no idea where to start…

Best Answer

Here are some hints.

  1. First it is not clear that $\mu$ is indeed a measure.
  2. Show that the sequence $\left( \int_{X}fd\mu_n\right)_{n\geqslant 1}$ is non-decreasing.
  3. As a consequence, $\lim_{n\to +\infty}\int_{X}fd\mu_n=\sup_{n\geqslant 1}\int_{X}fd\mu_n$.
  4. By definition of the Lebesgue integral, $$\int_{X}fd\mu_n=\sup_{s\in\mathcal S} \int_{X}sd\mu_n ,$$ where $\mathcal S=\left\{s\mid 0\leqslant s\leqslant f, .s=\sum_{i=1}^Nc_i\mathbf 1_{A_i},A_i\in\mathcal A\right\}$. Switching the two supremum reduces to establish the result when $f=\sum_{i=1}^Nc_i\mathbf 1_{A_i},A_i\in\mathcal A$. Indeed, by 3. $$ \lim_{n\to +\infty}\int_{X}fd\mu_n=\sup_{n\geqslant 1}\sup_{s\in\mathcal S}\int_{X}sd\mu_n=\sup_{s\in\mathcal S}\sup_{n\geqslant 1}\int_{X}sd\mu_n. $$ If we manage to show that for each $s\in\mathcal S$, the equality $\sup_{n\geqslant 1}\int_{X}sd\mu_n=\int_{X}sd\mu$, then we are done.
  5. By linearity it suffices to do it when $f =\mathbf 1_{A},A\in\mathcal A$.
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